\newproblem{lay:3_3_20}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 3.3.20}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Find the area of the parallelogram whose vertices are (-1,0), (0,5), (1,-4), (2,1).
}{
   % Solution
	Let us draw the parallelogram:
	Calling $\mathbf{x}_A=(-1,0)$, $\mathbf{x}_B=(0,5)$, $\mathbf{x}_C=(1,-4)$, the sought area is the absolute value of the determinant of the vectors $\mathbf{x}_B-\mathbf{x}_A$ and
	$\mathbf{x}_C-\mathbf{x}_A$.
	\begin{center}
		$\mathbf{x}_B-\mathbf{x}_A=\begin{pmatrix}0&5\end{pmatrix}-\begin{pmatrix}-1&0\end{pmatrix}=\begin{pmatrix}1&5\end{pmatrix}$\\
		$\mathbf{x}_C-\mathbf{x}_A=\begin{pmatrix}1&-4\end{pmatrix}-\begin{pmatrix}-1&0\end{pmatrix}=\begin{pmatrix}2&-4\end{pmatrix}$\\
		$\mathrm{abs}\left(\left|\begin{array}{cc} \mathbf{x}_B-\mathbf{x}_A & \mathbf{x}_C-\mathbf{x}_A \end{array}\right|\right)=
		 \mathrm{abs}\left(\left|\begin{array}{cc} 1 & 2 \\ 5 & -4\end{array}\right|\right)=14$
	\end{center}
}
\useproblem{lay:3_3_20}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
